(1) Which of the following statements are true and which are false?
Justify your answer.
(a) The product of two orthogonal n×n matrices is orthogonal.
Solution. True. Let A and B be two orthogonal matrices
of the same size. Then AT A = B T B = I. It follows
that (AB)T (AB) = B T (AT A)B = B T B = I. So AB is
orthogonal.
(b) Two matrices with the same characteristic polynomials are
similar.
Solution. False.
1
A=
0
Let
0
1 1
and B =
1
0 1
Then det(λI − A) = det(λI − B) = (λ − 1)2 . But A
and B are not similar. Otherwise, B = P −1 AP for an
invertible matrix P . But P −1 AP = P −1 P = I and B 6= I.
Contradiction.
(c) For two n × n invertible matrices A and B, AB and BA
are similar.
Solution. True since AB = B −1 (BA)B.
(d) rank(A) = rank(A2 ) for every square matrix A.
Solution. False. Let
0 1
A=
0 0
Then rank(A) = 1 but A2 = 0 and rank(A2 ) = 0.
(e) For a Householder matrix A and two vectors u and v, if
Au = v, then Av = u.
Solution. True. Since A is a Householder matrix, A is
symmetric and orthogonal. Therefore, A2 = AT A = I.
Consequently,
Au = v ⇒ A2 u = Av ⇒ u = Av
(f) If A is diagonalizable, so is AT .
Solution. True. Since A is diagonalizable, there is an
invertible matrix P such that P AP −1 = D is diagonal. It
follows that
DT = (P AP −1 )T = (P −1 )T AT P T = (P T )−1 AT P T
1
2
Since D = DT is diagonal, AT is diagonalizable.
(2) Show that two symmetric matrices are similar if and only if
they have the same characteristic polynomials.
Proof. Let A and B be two symmetric matrices.
Suppose that A and B are similar. Then A = P BP −1 for an
invertible matrix P . Then
det(λI − A) = det(λI − P BP −1 ) = det(P (λI − B)P −1 )
= det(P ) det(λI − B) det(P −1 ) = det(λI − B)
Therefore, A and B have the same characteristic polynomials.
On the other hand, suppose that A and B have the same
characteristic polynomials. Let
det(λI − A) = det(λI − B) = (λ − λ1 )(λ − λ2 )...(λ − λn )
where λ1 , λ2 , ..., λn are the eigenvalues of A and B.
Since A and B are symmetric, they are diagonalizable.
exist invertible matrices P and Q such that



λ1
λ1



λ
λ2
2
 and QBQ−1 = 
P AP −1 = 
.
..



..
.
λn
There




λn
Hence
P AP −1 = QBQ−1 ⇒ A = P −1 QBQ−1 P = (P −1 Q)B(P −1 Q)−1
So A and B are similar.
(3) Let u = (1, 1, 1) and v = (0, −1, −1) be two vectors in R3 .
(a) Find the hyperplane W = {a1 x1 + a2 x2 + a3 x3 = 0} such
that u is the reflection of v with respect to W .
Solution. We have W = w⊥ where w = u − v = (1, 2, 2).
So W = {x1 + 2x2 + 2x3 = 0}.
(b) Let T : R3 → R3 be the reflection with respect to the
hyperplane obtained in part (a). Find the matrix [T ]B
representing T with respect to the standard basis B.
3
Solution. The corresponding Householder matrix is
 
T
1 w
w
2
[T ]B = I − 2
= I − 2 1 2 2
||w|| ||w||
9 2

 

1 2 2
7/9 −4/9 −4/9
2
= I − 2 4 4 = −4/9 1/9 −8/9
9 2 4 4
−4/9 −8/9 1/9
(4) Let Q(x1 , x2 , x3 ) = (x1 + x2 + x3 )2 + (x1 − x2 )2 be a quadratic
form in the three variables x1 , x2 and x3 .
(a) Find a 3 × 3 symmetric matrix A such that
Q(x1 , x2 , x3 ) = xT Ax,
 
x1

where x = x2 .
x3
Solution. We have


2 0 1
Q(x1 , x2 , x3 ) = 2x21 + 2x22 + x23 + 2x1 x3 + 2x2 x3 = xT 0 2 1 x
1 1 1
(b) Find a 3 × 3 orthogonal matrix P such that P T AP is diagonal.
Solution. Note that (1, 1, 1) and (1, −1) are orthogonal.
Therefore,
√
√ 
 √
1/√3 1/ √3 1/ 3
P T = 1/ 2 −1/ 2
0 
a
b
c
Since P T is orthogonal, a+b+c = a−b = 0 and a2 +b2 +c2 =
1. Therefore,
√
√ 
 √
1/√3 1/ √3
1/ 3
P T = 1/√2 −1/√ 2
0√ 
1/ 6 1/ 6 −2/ 6
and hence
√
√ 
 √
1/√3 1/ √2
1/√6
P = 1/√3 −1/ 2 1/ √6 
1/ 3
0
−2/ 6
4
(c) Let x =P y. What is Q in terms of y1 , y2 and y3 , where
y1
y = y2 ?
y3
Solution. We have
Q(x1 , x2 , x3 ) = xT Ax = yT P T AP y = 3y12 + 2y22
(d) Find the maximum and minimum values of Q(x1 , x2 , x3 )
subject to the constraint x21 + x22 + x23 = 1 and where these
extreme values are achieved.
Solution. We have
0 ≤ 3y12 + 2y22 ≤ 3(y12 + y22 + y32 )
Therefore, Qmin = 0 when y1 = y2 = 0 and y3 = ±1, i.e.,
 √ 
 
 
1/√6
x1
0
x2  = P  0  = ±  1/ 6 
√
±1
x3
−2/ 6
and Qmax = 3 when y1 = ±1 and y2 = y3 = 0, i.e.,
 
 √ 
 
1/√3
±1
x1
x2  = P  0  = ± 1/ 3
√
0
x3
1/ 3
1 −1
.
(5) Let A =
−1 1
(a) Find an invertible matrix P such that P AP −1 is diagonal.
Solution. The characteristic polynomial of A is
det(λI − A) = λ(λ − 2).
For λ = 0, the corresponding eigenspace is
1
null(−A) = Span{
}
1
For λ = 1, the corresponding eigenspace is
1
null(2I − A) = Span{
}
−1
Let
−1
1 −1 −1
1/2 1/2
1 1
=
P =
=−
1/2 −1/2
1 −1
2 −1 1
5
Then
P AP
−1
1/2 1/2
1 −1 1 1
0 0
=
=
1/2 −1/2 −1 1
1 −1
0 2
(b) Find A2009 .
Solution. Since
PA
2009
P
−1
= (P AP
−1 2009
)
0
0
=
2009
0 2
we have
A
2009
0
0
=P
P
0 22009
1 1
0
0
1/2 1/2
=
1 −1 0 22009 1/2 −1/2
2008
2
−22008
=
−22008 22008
−1
Alternatively, we may apply the Cayley-Hamilton theorem:
A2 = 2A. It follows that An = 2n−1 A. Therefore,
2008
2
−22008
2009
2008
A
=2 A=
−22008 22008
(6) Let P2 be the vector space of polynomials in x of degree at most
2 and let T : P2 → P2 be the map given by
T (f (x)) = f (1 − x).
(a) Show that T is a linear transformation.
Proof. For all f (x) and g(x) in P2 ,
T (f (x)+g(x)) = (f +g)(1−x) = f (1−x)+g(1−x) = T (f (x))+T (g(x))
For all f (x) ∈ P2 and λ ∈ R,
T (λf (x)) = λf (1 − x) = λT (f (x))
Therefore, T is a linear transformation.
6
(b) Find the matrix [T ]S representing T with respect to the
basis S = {1, x, x2 }.
Solution. We have T (1) = 1, T (x) = 1 − x and T (x2 ) =
(1 − x)2 . Therefore,
 
 
1
1



[T (1)]S = [1]S = 0 , [T (x)] = [1 − x]S = −1
0
0
and


1
[T (x2 )]S = [(1 − x)2 ]S = [1 − 2x + x2 ]S = −2
1
It follows that


1 1
1
[T ]S = 0 −1 −2
0 0
1
(c) Let B = {1, x − 1, (x − 1)2 }. Find the transition matrix
PS→B and the matrix [T ]B representing T with respect to
B.
Solution. Since
 
 
1
1



[1]B = 0 , [x]B = [1 + (x − 1)]B = 1
0
0
and
 
1
[x2 ]B = [(1 + (x − 1))2 ]B = [1 + 2(x − 1) + (x − 1)2 ]B = 2
1
we have


1 1 1
[PS→B ] = 0 1 2
0 0 1
Since
 
 
1
−1
[T (1)]B = [1]B = 0 , [T (x − 1)] = [−x]B = [−1 − (x − 1)]B = −1
0
0
7
and
 
1
2
2
2

[T ((x − 1) )]B = [x ]B = [1 + 2(x − 1) + (x − 1) ]B = 2
1
we have


1 −1 1
[T ]B = 0 −1 2
0 0 1
(d) Does there exist a basis B such that [T ]B is diagonal? If
yes, find such a basis B. If no, justify your answer.
Solution. Such a basis exists if and only if [T ]S is diagonalizable. The matrix [T ]S has two eigenvalues 1 and −1.
For λ = 1, the corresponding eigenspace is
   
0
1
null(I − [T ]S ) = Span{0 ,  1 }
−1
0
For λ = −1, the corresponding eigenspace is
 
1

null(−I − [T ]S ) = Span{ −2}
0
Therefore, [T ]S has three linearly independent eigenvectors
and is hence diagonalizable. The corresponding basis B is
{1, x − x2 , 1 − 2x}.
(7) Do the following:
(a) Find the least squares solution of


 
1 0 8
1 1 x1 = 0
x2
1 0
0
Solution. The associated normal system is


 
1 0 8
1 1 1 
x
1
1
1
3
1
x1
8
1
0 ⇔
1 1
=
=
0 1 0
x2
0 1 0
1 1 x2
0
1 0
0
So the least squares solution is
x1 = 4
x2 = −4
8
(b) Express w = (8, 0, 0) in the form w = w1 + w2 , where
w1 lies in the subspace W of R3 spanned by the vectors
u1 = (1, 1, 1) and u2 = (0, 1, 0) and w2 is orthogonal to W .
Solution. Note that
w1 = ProjW w = Projcol(A) w
where


1 0
A = 1 1
1 0
By part (a),
w1 = x1 u1 + x2 u2 = (4, 0, 4)
and
w2 = w − w1 = (4, 0, −4)